1.4 Solvable Lie Algebras

Home , Nilpotent Lie algebra, Semisimple Lie algebra, Simple Lie algebra

1.4. SOLVABLE LIE ALGEBRAS 17

1.4 Solvable Lie algebras

1.4.1 Derived series and solvable Lie algebras The derived series of a Lie algebra L is given by: L(0) = L, L(1) = [L, L],,L(2) = [L(1),L(1)], ··· ,L(k) = [L(k−1),L(k−1)], ··· A Lie algebra L is called solvable if L(n) = 0 for some n. Ex. The Lie algebra t(n, F ) of n × n upper triangular matrices is solvable. (exercise) (k) Ex. Every nilpotent Lie algebra is solvable. (Prove L ⊂ L(k) for all k by induction on k.) We give some basic properties of solvable Lie algebras, and compare them with those of nilpotent Lie algebras. Prop 1.19. Let L be a Lie algebra. 1. If L is solvable, then so are all subalgebras and homomorphic images of L. (Similar for nilpotent case)

2. Given any ideal I E L, L is solvable iﬀ both I and L/I are solvable. (For nilpotent case, the “if” part only holds for special I, e.g. I = Z(L).) 3. If I and J are solvable ideals of L, then so is I + J. (Similar for nilpotent case. See ex 3.6 of Humphreys) Remark. Prop 1.19 implies that L has a unique maximal nilpotent solvable ideal, called the radical of L, denoted Rad L. L is called semisimple if Rad L = 0. Cor 1.20. Every Lie algebra L is decomposed as a solvable ideal Rad L and a semisimple homomorphic image L/Rad L: 0 → Rad L → L → L/Rad L → 0. Ex. A Lie algebra L is called simple if L has no ideals except itself and 0, and [L, L] 6= 0 (i.e. L is not abelian). Show that every simple Lie algebra is semisimple.

1.4.2 Lie’s Theorem In Engel’s theorem, we see that every nilpotent algebra L (dim L = n) has ad L isomorphic by conjugation to a subalgebra of n(n, F ) (of strictly upper triangular matrices). From now on, we always assume that F is algebraically closed and char F = 0, unless otherwise speciﬁed. Then Lie’s Theorem says that a solvable subalgebra of gl(n, F ) is isomorphic by conjugation to a subalgebra of t(n, F ) (of upper triangular matrices). Thm 1.21. Let L be a solvable subalgebra of gl(V ), 0 < dim V < ∞. Then V contains a common eigenvector for all the endomorphisms in L. Explicitly, there exist v ∈ V − {0} and λ ∈ L∗ such that x.v = λ(x)v for all x ∈ L. Remark. The claim is not true if F is not an algebraic closure. For example, F = R, and a b XY S = | a, b ∈ ,L = | X,Y,Z ∈ S . −b a R 0 Z x y Then L ' | x, y, z ∈ is solvable. However, L is not similar to a subalgebra of t(4, ). 0 z C R (why?) 18 CHAPTER 1. BASIC CONCEPTS

Proof of Theorem 1.21. We prove by induction on dim L. The case dim L = 0 is clear. We follow the steps below similar to the proof of Theorem 1.16:

1. Goal: ﬁnd a codimension one ideal K E L; then select z ∈ L − K, so that L = K + F z. Since L is solvable, we have L ) [L, L]. The quotient algebra L/[L, L] is abelian and every subspace is an ideal. The inverse image K of a codimension one subspace of L/[L, L] is a codimension one ideal in L.

2. Since K is solvable and dim K < dim L, by induction hypothesis, there is a common eigenvector v ∈ V for K. In other words, there is a linear functional λ : K → F , such that y.v = λ(y)v for any y ∈ K. So the common eigenspace of K for λ is nonzero:

W := {w ∈ V | y.w = λ(y)w, for all y ∈ K}.

3. Assume that W is L-invariant. Since F is algebraically closed, z ∈ L has an eigenvector v0 ∈ W . Since L = K + F z, v0 is a common eigenvector for L. We prove the theorem. 4. Remaining goal: Prove that W is L-invariant. Fix x ∈ L and w ∈ W . For any y ∈ K and any integer i > 0,

y(x.w) = xy.w − [x, y].w = λ(y)x.w − λ([x, y])w. (1.2)

To let x.w ∈ W , we have to prove that λ([x, y]) = 0. Let n > 0 be the smallest integer for which w, x.w, ··· , xn.w is linearly dependent. Deﬁne

i−1 Wi = span{w, x.w, ··· , x .w}.

Then dim Wi = i for i < n, and dim Wn = dim Wn+1 = ··· = n. i i We prove that y(x .w) ∈ λ(y)x .w + Wi for any y ∈ K and any integer i > 0 by induction i−1 i−1 on i. The case i = 1 is done by (1.2). Now suppose y(x .w) = λ(y)x .w + wi−1, where wi−1 ∈ Wi−1. Then

yxi.w = x(yxi−1.w) − [x, y]xi−1.w i−1 i−1 = x(λ(y)x .w + wi−1 ) − [x, y]x .w (by induction hypothesis) |{z} | {z } in Wi−1 in Wi i ∈ λ(y)x .w + Wi.

The claim is proved.

Now for any y ∈ K, all of x, y, and [x, y] ∈ K stabilize Wn. Relative to the basis n−1 w, x.w, ··· , x .w of Wn,[x, y] is represented by an upper triangular entries with λ([x, y])

as diagonal entries. Therefore, 0 = TrWn ([x, y]) = nλ([x, y]). Since char F = 0, we have λ([x, y]) = 0 as desired.

Cor 1.22 (Lie’s Theorem). Let L be a solvable subalgebra of gl(V ), dim V < ∞. Then L stabilizes some flag in V . In other words, the matrices of L relative to a suitable basis of V are upper triangular. Remark. Any finite dimension representation of a solvable algebra L, φ : L → gl(V ), implies that φ(L) is solvable and thus stabilizes a flag of V . 1.4. SOLVABLE LIE ALGEBRAS 19

Cor 1.23. Let L be solvable. Then there exists a chain of ideals of L, 0 = L0 ⊂ L1 ⊂ · · · ⊂ Ln = L, such that dim Li = i. Proof. Apply Lie’s Theorem to adjoint representation ad : L → gl(L).

Cor 1.24. Let L be solvable. Then x ∈ [L, L] implies that ad Lx is nilpotent. In particular, [L, L] is nilpotent. Proof. ad L ⊂ gl(L) is solvable. So ad L consists of upper triangular matrices relative to a suitable basis of L. Then ad [L, L] = [ad L, ad L] consists of strictly upper triagular matrices, which are nilpotent.

1.4.3 Jordan-Chevalley decomposition

We explore some linear algebra results in this subsection. Let Jn(λ) denote the n × n Jordan block with diagonal entries λ, and Jn := Jn(0) denote the n × n matrix that has 1 on the superdiagonal and 0 elsewhere.

Ex. 1. Jn(λ) = λIn + Jn, where λIn is diagonal, Jn is nilpotent, and λIn and Jn are commutative;

2. A Jordan matrix x = Jn1 (λ1) ⊕ Jn2 (λ2) ⊕ · · · can be decomposed as x = xs + xn, where xs is diagonal, xn is nilpotent, xs and xn are commutative;

−1 3. Given the decomposition of Jordan matrix x = xs + xn above, any similar matrix gxg of x can be decomposed as −1 −1 −1 gxg = gxsg + gxng , −1 −1 where gxsg is diagonalizable, gxng is nilpotent, and they commute to each other. A matrix x ∈ F n×n (or x ∈ End (V )) is called semisimple if the roots of its minimal polynomial over F are all distinct. Equivalently, when F is algebraically closed, x is semisimple iff x is diagonalizable. Prop 1.25. Let V be a finite dimensional vector space over an arbitrary field F , and x ∈ End (V ).

1. There exist unique xs, xn ∈ End (V ) satisfying the (additive) Jordan-Chevalley decomposition: x = xs + xn, where xs is semisimple, xn is nilpotent, xs and xn commute.

2. There exist polynomials p(T ) and q(T ) without constant term, such that xs = p(x) and xn = q(x). In particular, xs and xn commute with any endomorphism commuting with x. Proof for complex case: By choosing an appropriate base of V , we may assume that the matrix form of x is in Jordan canonical form. Then the decomposition x = xs + xn is explicitly constructed in the proceeding example; moreover, x, xs and xn commute. It remains to show that xs and xn are polynomials expressions of x.

If x has only one eigenvalue λ, then x = Jn1 (λ)⊕Jn2 (λ)⊕· · ·⊕Jnk (λ) has the Jordan-Chevalley n decomposition x = λ + xn where xn = x − λ. Since (x − λ) = 0, we can express λ as a polynomial of x without constant term, say λ = p(x), then xn = x − p(x) is also expressed as a polynomial of x without constant term. Now suppose x is a direct sum of Jordan matrices x(1) ⊕ · · · ⊕ x(t) corresponding to distinct (i) m (i) eigenvalues λ1, ··· , λt, each Jordan matrix x has the minimal polynomial (z − λi) i , and x = (i) (i) (i) (i) (i) (i) xs + xn where xs = pi(x ) and xn = qi(x ). The minimal polynomial of x is h(z) = 20 CHAPTER 1. BASIC CONCEPTS

Qt mi mi i=1(z − λi) . The polynomial ring C[z] is PID. For each i ∈ [t], the polynomials (z − λi) and mi Q mj h(z)/(z − λi) = j6=i(z − λj) are relatively prime. By Chinese Remainder Theorem, there exists ri(z) ∈ C[z] with degree less than deg h(z), such that

mi Y mj ri(z) ≡ 1 mod (z − λi) , ri(z) ≡ 0 mod (z − λj) . j6=i

(i) (j) It implies that ri(x ) = 1 and ri(x ) = 0 for j 6= i. Therefore, let

t t X X p(z) = pi(z)ri(z), q(z) = qi(z)ri(z). i=1 i=1

Then xs = p(x) and xn = q(x) as desired. Remark. The Jordan decomposition exists in any ﬁeld (See textbook, or we can modify the above proof to deal with arbitrary ﬁeld.) Moreover, it is relative to the abstract Jordan decomposition for elements on any semisimple Lie algebra (later section).

Lem 1.26. Let x ∈ End V , dim V < ∞, x = xs + xn its Jordan decomposition. Then ad x = ad xs + ad xn is the Jordan decomposition of ad x (in End (End V ) ) Proof. Suppose dim V = n, and x is in Jordan canonical form for a given basis of V . Then {eij | i, j ∈ [n]} is a basis of End V in which ad xs is diagonal and ad xn is nilpotent. Remark. The Jordan decomposition is indeed preserved by any ﬁnite dimensional representation ρ : End V → End W , that is, ρ(x) = ρ(xs) + ρ(xn) is the Jordan decomposition of ρ(x) in End W . Lem 1.27. Let U be a ﬁnite dimensional F -algebra. Then Der U contains the semisimple and nilpotent parts (in End U) of all its elements.

Proof. Suppose δ ∈ Der U has the Jordan decomposition δ = σ + ν in End U. For a ∈ F , we set

k Ua = {x ∈ U | (δ − a · 1) x = 0 for some k}. a Then U = Ua where spec(δ) = spec(σ) denotes the spectrum of δ (or σ), and σ acts on Ua a∈spec(δ) as a scalar multiplication by a. For any n ∈ Z+, by induction (exercise), the derivation δ satisﬁes that n X n (δ − (a + b) · 1)n(xy) = ((δ − a · 1)n−ix) · ((δ − b · 1)iy). i i=0

For any x ∈ Ua and y ∈ Ub, we have xy ∈ Ua+b. Therefore, σ(xy) = (a + b)xy = σ(x)y + xσ(y). Therefore, σ is a derivation. So σ, ν ∈ Der U.

1.4.4 Cartan’s Criterion for Solvability Lie’s Theorem says that a linear Lie algebra L ∈ gl(V ) is solvable iﬀ it is a subalgebra of t(n, F ) relative to a suitable basis. Then [L, L] consists of strictly upper triangular matrices, and Tr (xy) = 0 for all x ∈ L and y ∈ [L, L]. It turns out that the converse is also true.

Thm 1.28 (Cartan’s Criterion). Let L be a subalgebra of gl(V ), V ﬁnite dimensional. Then L is solvable iﬀ Tr (xy) = 0 for all x ∈ [L, L] and y ∈ L. 1.4. SOLVABLE LIE ALGEBRAS 21

Proof for “⇐=” in complex case: We prove that if Tr (xy) = 0 for any x ∈ [L, L] and y ∈ L, then ad x is nilpotent. By Engel’s Theorem, it implies that [L, L] is nilpotent, so that L is solvable. Let M = {z ∈ gl(V ) | [z, L] = ad z(L) ⊆ [L, L]}. Clearly L ⊆ M. Let [x, y] be a typical generator of [L, L], and z ∈ M. By computation (exercise),

Tr ([x, y]z) = Tr (x[y, z]) = 0, by [y, z] ∈ [L, L] and the hypothesis. So Tr (xz) = 0 for any x ∈ [L, L] and z ∈ M. Fix x ∈ [L, L]. Let x = s + n be the Jordan decomposition of x. We show that s = 0, so that ad x = ad n is nilpotent. Now ad x = ad s + ad n is the Jordan decomposition of ad s in End L, and ad s is a polynomial of ad x without constant term. Since ad x(L) ⊆ [L, L], we have ad s(L) ⊆ [L, L] so that s ∈ M. With an appropriate basis of V , s has the matrix form diag (a1, ··· , am). Let {eij | i, j ∈ [m]} be the standard basis of gl(V ). Then ad s(eij) = (ai − aj)eij. So ad s is diagonal w.r.t. the basis {eij}. By Lagrange interpolation, there exists a polynomial r(T ) ∈ C[T ] without constant term, such that 1 r(ai − aj) = ai − aj. Then ad s = r(ad s), which implies that ad s(L) ⊆ [L, L]. Therefore, s ∈ M, so that m X 2 0 = Tr (xs) = Tr (ss) = |ai| . i=1 We get s = 0 as desired.

Cor 1.29. A Lie algebra L is solvable iﬀ Tr (ad x ad y) = 0 for all x ∈ [L, L], y ∈ L.

We will recall this corollary in the next section about Killing form.

1This part shoud be modiﬁed for the other alg closed char 0 ﬁelds F . See Humphrey’s text.